126 divide 6' long division

Mathematics

1 answer:

swat323 years ago

3 0

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Convert the integral below to polar coordinates and evaluate the integral.

White raven [17]

I suppose the integral is

$\displaystyle\int_0^5\int_y^{\sqrt{25-y^2}} xy\,\mathrm dx\,\mathrm dy$

The integration region corresponds to a sector of a cirlce with radius 5 subtended by a central angle of π/4 rad. We can capture this region in polar coordinates by the set

$R=\left\{(r,\theta)\mid0\le r\le5,0\le\theta\le\dfrac\pi4\right\}$

Then $x=r\cos\theta$ , $y=r\sin\theta$ , and $\mathrm dx\,\mathrm dy=r\,\mathrm dr\,\mathrm d\theta$ . So the integral becomes

$\displaystyle\iint_Rxy\,\mathrm dx\,\mathrm dy=\int_0^{\pi/4}\int_0^5r^3\sin\theta\cos\theta\,\mathrm dr\,\mathrm d\theta=\frac{625}{16}$

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A function is given by the equation h (x)=x 2 squared +6x-3. what is h (3)?

Leno4ka [110]

Your "<span>h (x)=x 2 squared +6x-3" is ambiguous. If you meant "x squared," then you need only write x^2 OR "x squared," but NOT "x 2 squared."

I will assume that by "</span><span>h (x)=x 2 squared +6x-3" you actually meant:

</span><span>h(x)=x^2 +6x-3. To find h(3), subst. 3 for x: h(3) = (3)^2 + 6(3) - 3, or

h(3) = 9 + 18 - 3, or h(3) = 24.</span>

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stepan [7]

So, what’s the question, I probably can answer it if you include the question

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